Problem: The equation of a circle $C$ is $x^2+y^2-8x+12y+48 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-8x) + (y^2+12y) = -48$ $(x^2-8x+16) + (y^2+12y+36) = -48 + 16 + 36$ $(x-4)^{2} + (y+6)^{2} = 4 = 2^2$ Thus, $(h, k) = (4, -6)$ and $r = 2$.